Question

Prove that, if two lines containing chords of a circle intersect each other outside the circle,
then the measure of angle between them is half the difference in measures of the arcs
intercepted by the angle.​

Answer 1

Answer:

Given: Chord AB and chord CD intersect at E in the exterior of the circle. To prove: ∠AEC = 1/2 [m(arc AC) – m(arc BD)] Construction: Draw seg AD. Proof: ∠ADC is the exterior angle of ∆ADE. ∴ ∠ADC = ∠DAE + ∠AED [Remote interior angle theorem] ∴ ∠ADC = ∠DAE + ∠AEC [C – D – E] ∴ ∠AEC = ∠ADC – ∠DAE ……(i) ∠ADC = 1/2m(arc AC) (ii) [Inscribed angle therom] ∠DAE = 1/2 m(arc BD) (iii) [A – B – E, Inscribed angle theorem] ∴ ∠AEC = 1/2 m(arc AC) – 1/2 m (arc BD) [From (i), (ii) and (iii)] ∴ ∠AEC = 1/2 m(arc AC) – m (arc BD)Read more on Sarthaks.com - https://www.sarthaks.com/851639/prove-lines-containing-chords-circle-intersect-other-outside-circle-measure-angle-between